A) \[\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)\text{t}\]
B) \[\left( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)\]
C) \[\frac{\left( {{\alpha }^{{}}}+{{\beta }^{{}}} \right)\text{t}}{\alpha \beta }\]
D) \[\frac{\alpha \beta \text{t}}{{{\alpha }^{{}}}+{{\beta }^{{}}}}\]
Correct Answer: D
Solution :
In fig., \[\text{A}\text{.}{{\text{A}}_{\text{1}}}\text{=}{{\text{v}}_{\text{max}\text{.}}}=\alpha {{\text{t}}_{\text{1}}}=\beta {{\text{t}}_{\text{2}}}\] \[\text{But t=}{{\text{t}}_{1}}+{{\text{t}}_{2}}=\frac{{{V}_{\max }}}{\alpha }+\frac{{{V}_{\max }}}{\beta }\] \[={{\text{v}}_{\max }}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)={{\text{v}}_{\max }}\left( \frac{\alpha +\beta }{\alpha \beta } \right)\] \[or,\text{ }{{\text{v}}_{\max }}=t\left( \frac{\alpha \beta }{\alpha +\beta } \right)\]You need to login to perform this action.
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